Problem: Solve for $x$ : $ 5|x - 3| + 2 = -6|x - 3| + 8 $
Solution: Add $ {6|x - 3|} $ to both sides: $ \begin{eqnarray} 5|x - 3| + 2 &=& -6|x - 3| + 8 \\ \\ { + 6|x - 3|} && { + 6|x - 3|} \\ \\ 11|x - 3| + 2 &=& 8 \end{eqnarray} $ Subtract ${2}$ from both sides: $ \begin{eqnarray} 11|x - 3| + 2 &=& 8 \\ \\ { - 2} &=& { - 2} \\ \\ 11|x - 3| &=& 6 \end{eqnarray} $ Divide both sides by ${11}$ $ \dfrac{11|x - 3|} {{11}} = \dfrac{6} {{11}} $ Simplify: $ |x - 3| = \dfrac{6}{11}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{6}{11} $ or $ x - 3 = \dfrac{6}{11} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{6}{11} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{6}{11} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{6}{11} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $11$ $ x = - \dfrac{6}{11} {+ \dfrac{33}{11}} $ $ x = \dfrac{27}{11} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{6}{11} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{6}{11} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{6}{11} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $11$ $ x = \dfrac{6}{11} {+ \dfrac{33}{11}} $ $ x = \dfrac{39}{11} $ Thus, the correct answer is $x = \dfrac{27}{11} $ or $x = \dfrac{39}{11} $.